Calling ShowDialog in BackgroundWorker

Question

I have a WinForms application in which my background worker is doing a sync task, adding new files, removing old ones etc.

In my background worker code I want to sho

Answer 1

IMO answers stating that you should launch a thread to handle this are misguided. What you need is to jump the window back to the main dispatcher thread.

In WPF

public ShellViewModel(
    [NotNull] IWindowManager windows, 
    [NotNull] IWindsorContainer container)
{
    if (windows == null) throw new ArgumentNullException("windows");
    if (container == null) throw new ArgumentNullException("container");
    _windows = windows;
    _container = container;
    UIDispatcher = Dispatcher.CurrentDispatcher; // not for WinForms
}

public Dispatcher UIDispatcher { get; private set; }

and then, when some event occurs on another thread (thread pool thread in this case):

public void Consume(ImageFound message)
{
    var model = _container.Resolve();
    model.ForImage(message);
    UIDispatcher.BeginInvoke(new Action(() => _windows.ShowWindow(model)));
}

WinForms equivalent

Don't set UIDispatcher to anything, then you can do have:

public void Consume(ImageFound message)
{
    var model = _container.Resolve();
    model.ForImage(message);
    this.Invoke( () => _windows.ShowWindow(model) );
}

DRYing it up for WPF:

Man, so much code...

public interface ThreadedViewModel
    : IConsumer
{
    /// 

    /// Gets the UI-thread dispatcher
    /// 
    Dispatcher UIDispatcher { get; }
}

public static class ThreadedViewModelEx
{
    public static void BeginInvoke([NotNull] this ThreadedViewModel viewModel, [NotNull] Action action)
    {
        if (viewModel == null) throw new ArgumentNullException("viewModel");
        if (action == null) throw new ArgumentNullException("action");
        if (viewModel.UIDispatcher.CheckAccess()) action();
        else viewModel.UIDispatcher.BeginInvoke(action);
    }
}

and in the view model:

    public void Consume(ImageFound message)
    {
        var model = _container.Resolve();
        model.ForImage(message);
        this.BeginInvoke(() => _windows.ShowWindow(model));
    }

Hope it helps.